int(1)/((x+1)(x-2))dx=A log_(e)(x+1)+8log(x-2)+c

int (dx)/((x+1)(x-2))=A log (x+1)+B log (x-2)+C , where

int (dx)/((x+1)(x-2))=A log (x+1)+B log (x-2)+C , where

int dx/((x+1)(x-2)) = A log (x + 1) + B log (x - 2) + c where

int dx/((x+1)(x+2)) = A log (x + 1) + B log (x +2) + c where

int(1)/(1+e^(x))dx= (a) log(1+e^(x)) (b) log((1+e^(x))/(e^(x))) (c) log(1+e^(-x)) (d) -log(e^(-x)+1)

int(1)/(x cos^(2)(log_(e)x))dx

int(1)/(x cos^(2)(log_(e)x))dx

If intx log (1+(1)/(x))dx =f(x).log_(e)(x+1)+g(x)log_(e)x^(2)xLx+C , then

int x log_(e)(1+x)dx

int(log(x+1)-log x)/(x(x+1))dx= (A) log(x-1)log x+(1)/(2)(log x-1)^(2)-(1)/(2)(log x)^(2)+c (B) (1)/(2)(log(x+1))^(2)+(1)/(2)(log x)^(2)-log(x+1)log x+c (C) -(1)/(2)(log(x+1)^(2))-(1)/(2)(log x)^(2)+log x*log(x+1)+c (D) [log(1+(1)/(x))]^(2)+c